Can a meager linear subspace be written as a countable increasing union of nowhere dense subspaces?

Let $X$ be a separable Banach space. In this question, "subspace" means a linear subspace, not necessarily closed.

Suppose $E \subset X$ is a subspace which is meager, so that we can write $E = \bigcup_n E_n$, where the $E_n$ are nowhere dense subsets of $X$. Without loss of generality, we can also take $E_1 \subset E_2 \subset \cdots$. Can the $E_n$ be taken to be subspaces of $X$? That is, can a meager subspace always be written as a countable increasing union of nowhere dense subspaces?

Note that the linear span of a nowhere dense set is not necessarily nowhere dense (consider the unit sphere), nor is the sum of two nowhere dense subspaces (for instance, in $C([0,1])$, consider the mean-zero functions and the constants).

This came up while thinking about this answer: it isn't sufficient to check weak(-*) convergence in $L^2$ on the subspace $C([0,1])$; a sequence of linear functionals may converge pointwise on $C([0,1])$ but diverge somewhere else. More generally, when is it sufficient to check weak-* convergence on a dense subspace $A \subset X$? It is sufficient if $A$ is nonmeager, by a version of the uniform boundedness principle and a triangle inequality argument. But a subspace which is nonmeager lacks the Baire property, so such an $A$ would be a "weird" subspace, something we are not likely to encounter in everyday life. Most of the examples of dense subspaces I know are either countable dimension (hence meager), or complete in a stronger norm (hence analytic in $X$, hence has the BP, hence meager; as for example $C([0,1]) \subset L^2([0,1])$).

So I was wondering whether one could prove that it is never sufficient to check weak-* convergence on a meager subspace. If my question above has an affirmative answer, then we can do the following: for any meager subspace $E$, write $E = \bigcup_n E_n$ where $E_n$ are increasing nowhere dense subspaces. Since $E_n$ is nowhere dense, it is not dense, so by Hahn-Banach we may find $f_n \in X^*$ with $f_n(E_n) = 0$ and $\|f_n\| = n$. Then $f_n(x) \to 0$ for every $x \in E$, but $\{f_n\}$ is unbounded so (by the uniform boundedness principle) it is not weak-* convergent. Thus it would not suffice to check weak-* convergence on $E$.

Let $X = C_0([0,1])$, the space of continuous functions $\omega$ on $[0,1]$ having $\omega(0)=0$. Fix $0 < \alpha < 1/2$ and let $E = C^{0,\alpha}([0,1]) \cap X$ be the subspace of $\alpha$-Hölder continuous functions. Then $E$ is meager in $X$: By Arzelà-Ascoli, the closed balls of the Hölder norm are compact in $X$, hence by Riesz's lemma they are nowhere dense, and $E$ is their countable union.
Now let $\mu$ be the Wiener measure on $X$. It is well known that $\mu(E) = 1$, since Brownian motion is almost surely $\alpha$-Hölder continuous for any $\alpha < 1/2$. On the other hand, suppose $F$ is any nowhere dense subspace of $X$, so that its closure is a proper closed subspace of $X$. By Hahn-Banach there is a nonzero continuous linear functional $f$ that vanishes on $F$. Now $\mu$ is a nondegenerate Gaussian measure on $X$, so $f$ has a nondegenerate one-dimensional Gaussian distribution under $\mu$; in particular, $\mu(F) \le \mu(\{f = 0\}) = 0$. Thus every nowhere dense subspace has measure zero, so by countable additivity, $E$ cannot be a countable union of such.